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Proof step 15 from proof in-list-put (conclusion)

⊦ ¬ rk .key ∈ sks ∧ sks ≠ [] → (r -- sks ∈ st[rk, val] ↔ r -- sks ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ sks = [], r -- sks ∈ st[rk, val] ↔ r -- sks ∈ st, rk .key ∈ sks

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]